dijkstra's algorithm

Dijkstra Algorithm: Finding shortest paths in order

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Find shortest paths from source s to all other destinations

Djikstra's algorithm (named after its discover, E.W. Dijkstra) solves the problem of finding the shortest path from a point in a graph (the source) to a destination.

• Suppose we want to find a shortest path from a given node a to other nodes in a network (one-to-all shortest path problem)

• It finds the shortest path from a given node a to all other nodes in the network

• Node a is called a starting node or an initial node

we will find the shortest path from a to e.

Dijkstra's algorithm keeps two sets of vertices:

Set P : The set of vertices whose shortest paths from the source have already been determined andSet T : The remaining vertices.

So here initiallyP = {} i.e. EmptyT = {a,b,c,d,f}

Step 1:Label a with 0 and all others with .Labels are shortest paths from a to vertices. So we have P = { a }T = { b, c, d, e} L(a) = 0 L(b) = L(c) = L(d) = L(e) =

Step 2:• Nodes b, c, and d can be reached from the

current node a • Therefore by using L( i ) = min { L( n ), L( j ) + w( i , j ) }Where L(i) = label we are Updating. L(j) = Current Label Update distance values for these nodes

So we get L ( b ) = min { L( b ) , L( a ) + W ( a , b ) } = min { , 0 + 9 } = min { , 9 } Selecting the minimum we get L (b) = 9Smilarly calculating for L(c) and L(d) we get L(c) = min { , 19} = 19 L(d) = min { , 25} = 25

• Now, among the nodes b, c, and d, node b has the smallest distance value.

• So the status label of node b changes to permanent, while the status of c and d remains temporary

Step 3:• P = { a, b}• T = { c, d, e}• Node b becomes the current node Operation

• Nodes c, and e can be reached from the current node b. So, updating values for c, d, e

We get• L (c)= min { 19 , 9+16 } = min { 19 , 25} = 19• Similarly ( e ) = 45

• Now label c has the smallest value. Therefore it changes to permanent.

Step 4:• P = {a , b, c}• T = { d, e }• Updating labels for d and e, we get• L (d) = 24 and L (e) = 50

• Step 5 :• P ={ a,b,c,d } , T = {e}• Updating label e , we get L (e) = 45• So 45 is the shortest value which can be travelled to

reach node e.

Applications

Telephone NetworkThe vertices represents switching station, the edges represents the transmission line, and the weight of edges represents the BW.

Flight Agenda To determine the earliest arrival time for the destination given an origin airport and start time.